Question: Let $h$ be a twice differentiable function, and let $h(-2)=2$, $h'(-2)=0$, and $h''(-2)=-1$. What occurs in the graph of $h$ at the point $(-2,2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-2,2)$ is a minimum point. (Choice B) B $(-2,2)$ is a maximum point. (Choice C) C There's not enough information to tell.
Explanation: Since $h'(-2)=0$, we know that $x=-2$ is a critical point. The second derivative test allows us to analyze what happens in the graph of $h$ at this point according to these three cases: If $h''(-2)>0$, the graph of $h$ has a minimum point at $x=-2$. If $h''(-2)<0$, the graph of $h$ has a maximum point at $x=-2$. If $h''(-2)=0$, the test is inconclusive. [Why is this so?] We are given that $h''(-2)=-1<0$. Therefore, $(-2,2)$ is a maximum point.